Two capacitors of capacitances 20.0 pF and 50.0 pF are connected in series with a 6.00 V battery. Find (a) the potential difference across each capacitor and (b) the energy stored in each capacitor.

Question

Kevin Nash · Answer

Sol. C base 1 = 20 PF = 20 × 10^–12 F, C base 2 = 50 PF = 50 × 10^–12 F Effective C = C base 1C base 2/C base 1 + C base 2 = 2 * 10^-11 * 5 * 10^-11/2 * 10^-11 + 5 * 10^-11 = 1.428 * 10^-11 F Charge ‘q’ = 1.428 * 10^-11 * 6 = 8.568 * 10^-11 C V base 1 = q/C base 1 = 8.568 * 10^-11/2 * 10^-11 = 4.284 V V base 2 = q/C base 2 = 8.568 * 10^-11/5 * 10^-11 = 1.71 V Energy stored in each capacitor E base 1 = (1/2) C base 1V base 1^2 = (1/2) * 2 * 10^-11 * (4.284)^2 = 18.35 * 10^-11 = 184 PJ E base 2 = (1/2) C base 2V base 2^2 = (1/2) * 5 * 10^-11 * (1.771)^2 = 7.35 * 10^-11 = 73.5 PJ

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Two capacitors of capacitances 20.0 pF and 50.0 pF are connected in series with a 6.00 V battery. Find (a) the potential difference across each capacitor and (b) the energy stored in each capacitor.

Two capacitors of capacitances 20.0 pF and 50.0 pF are connected in series with a 6.00 V battery. Find (a) the potential difference across each capacitor and (b) the energy stored in each capacitor.

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